Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Calculate of <img alt="\mathrm{\mathrm{NH}_4 \mathrm{OH}}" src="https://entra

Calculate \mathrm{\Lambda_m^0} of \mathrm{\mathrm{NH}_4 \mathrm{OH}} in \mathrm{\ scm^{2}\ \mathrm{mol}^{-1}} and percentage decomposition of \mathrm{0.05\ \mathrm{M}\ \ \mathrm{NH}_4 \mathrm{OH}\ {\text {with }}\ \Lambda_m=25.1\ \mathrm{scm}^2 \mathrm{~mol}^{-1}} at a temperature of \mathrm{298 \mathrm{~K}} where at infinite dilution, molar conductivities \mathrm{\left(\Lambda_m^{\circ}\right)} of \mathrm{\mathrm{NH}_4 \mathrm{Cl},\ \mathrm{KOH}} and \mathrm{KCl} are \mathrm{155.3,\ 275.8} and \mathrm{150.8\ \mathrm{scm}^2 \mathrm{mol}^{-1}}, respectively.

Option: 1

275.6, \ 0.91%


Option: 2

30,\ 84


Option: 3

280.3,\ 8.9\%


Option: 4

280.3,\ 0.089


Answers (1)

best_answer

\mathrm{\Lambda_{m N H_4 O H}^{\circ}=\Lambda_m^{\circ}\left(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{KOH}\right)-\Lambda_{m K \mathrm{Cl}}^{\circ}}
\mathrm{\begin{aligned} & =155.3+275.8-150.8 \\ & =280.3\ \mathrm{scm}^2 \mathrm{~mol}^{-1} \end{aligned}}
Degree of dissociation 
\mathrm{\begin{aligned} \alpha & =\frac{\Lambda_m}{\Lambda_m^0} \\\ & =\frac{25.1}{280.3}=0.0895 \end{aligned}} \\\\ \mathrm{\%\ of\ degree\ of\ dissociation\ =8.9 \%}

Posted by

Rishi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question